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Question 52
If $$\frac{6x}{(2x^2 + 5x - 2)} = 1, x > 0,$$ then the value of $$x^3 + \frac{1}{x^3}$$ is:
Solution
We have $$\frac{6x}{2x^2-+5x-2}=1$$
Dividing numerator and denominator by x
we get $$\frac{6}{2x-\frac{2}{x}+5}=1$$
we get 2x-2/x =1
x-1/x =1/2
Now taking square we get
x^2+1/x^2-2 =1/4
we get x^2+1/x^2=9/4
we get (x+1/x)^2-2 =9/4
we get x+1/x =$$\frac{\sqrt{\ 17}}{2}$$
Taking cube we get
x^3+1/x^3+3(x+1/x) =$$\frac{17\sqrt{\ 17}}{8}$$
solving we get x^3+1/x^3=$$\frac{5\sqrt{\ 17}}{8}$$
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